1. How far in front of a lens of 45 cm focal length must an object be placed if the image is to be three times as big as the object?
(2)
2. Explain how a converging lens can be used to project an image over a large distance such as
in a cinema. Explain why getting the image in focus can be difficult.
(3)​

Lenses are found in a huge array of optical instruments, ranging from a simple magnifying glass to the eye to a camera’s zoom lens. In this section, we will use the law of refraction to explore the properties of lenses and how they form images.

The word lens derives from the Latin word for a lentil bean, the shape of which is similar to the convex lens in Figure 1. The convex lens shown has been shaped so that all light rays that enter it parallel to its axis cross one another at a single point on the opposite side of the lens. (The axis is defined to be a line normal to the lens at its center, as shown in Figure 1.) Such a lens is called a converging (or convex) lens for the converging effect it has on light rays. An expanded view of the path of one ray through the lens is shown, to illustrate how the ray changes direction both as it enters and as it leaves the lens. Since the index of refraction of the lens is greater than that of air, the ray moves towards the perpendicular as it enters and away from the perpendicular as it leaves. (This is in accordance with the law of refraction.) Due to the lens’s shape, light is thus bent toward the axis at both surfaces. The point at which the rays cross is defined to be the focal point F of the lens. The distance from the center of the lens to its focal point is defined to be the focal length f of the lens. Figure 2 shows how a converging lens, such as that in a magnifying glass, can converge the nearly parallel light rays from the sun to a small spot.

Explanation:

For convex lens sign convention is-

V=+ve

a=−ve

f=+ve

I (length of image) =−ve (real image)

O (size of object) =+ve

So, using lens formula −  +O −I =  u +V ⇒u=−  3 v ⇒v=−3u v 1  −  u 1  =  f 1 ⇒  −3u 1  −  u 1 =  f 1 ⇒  3u 4 =−  15 1 ⇒u=−20cm  object is placed 20cm in front of the lens

i was thinking more like reflects green light, just not sure

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation: